Given,

Mass, $m=2 \text{kg}$,

Spring constant, $k=2\times {10}^4 \mathrm{N} {\mathrm{m}}^{-1}$

Acceleration due to gravity, $g=10 \mathrm{m} {\mathrm{s}}^{-2}$.

Applying Law of conservation of mechanical energy,

$mg \left(h+x\right)=\frac{1}{2} k{x}^2$

$\Rightarrow k{x}^2-2mgx-2mgh=0$

$\Rightarrow \left(2\times {10}^4\right)x^2-40x-48=0$

Using Quadratic formula,

$\Rightarrow x=\frac{40\pm \sqrt{{\left(-40\right)}^2-4\times 2\times {10}^4\times \left(-48\right)}}{2\times 2\times {10}^4}$

$\Rightarrow x=\frac{40\pm 1960}{4\times {10}^4}$

$x$ cannot be negative, thus

$x=\frac{40+1960}{4\times {10}^4}=50 \mathrm{mm}$.